Can we connect these two matrices with a piecewise linear path in $GL_n(\mathbb R)$?

Question

Suppose $A\in GL_n(\mathbb R)$ with all eigenvalues lying on the open right half plane, i.e., with positive real part. Let $B = S^{-1} A S$ where $S \in GL_n(\mathbb R)$. I want to define a continuous path $\gamma(t): [0,1] \to GL_n(\mathbb R)$ with $\gamma(0) = A$ and $\gamma(1) = B$. This path certainly exists since $\det(A) = \det(B) > 0$ and $GL_n(\mathbb R)_+$ is connected. I am wondering whether we can make the path $\gamma$ to be piecewise linear.

Answer 1

Here is a concrete construction. Since $SL_n(\mathbb R)$ is generated by elementary matrices of the shear type, $P:=\det(A)^{-1/n} A$ can be written as a product $E_1E_2\cdots E_m$ of such shear matrices. Hence $A$ is connected to the identity matrix via a piecewise linear path on $GL_n^+(\mathbb R)$: $$ A\ \to\ P = E_1E_2\cdots E_m\ \to\ E_1E_2\cdots E_{m-1}\ \to\ \cdots\to\ E_1E_2\to\ E_1\ \to\ I. $$

Answer 2

Yes. Consider any continuous path $\gamma: [0,1] \to GL_n(\mathbb R)$. Since $\gamma([0,1])$ is compact and $GL_n(\mathbb R)$ is open in $\mathbb R^{n\times n}$, given a norm on $\mathbb R^{n\times n}$ there is some $\epsilon > 0$ such that all matrices within distance $\epsilon$ of $\gamma([0,1])$ are in $GL_n(\mathbb R)$. Take $t_0 = 0 < t_1 < \ldots < t_n = 1$ so that $\|\gamma(t_j) - \gamma(t_{j+1}\| < 2 \epsilon$ for $j=0 \ldots n-1$. Then the piecewise linear function $\tilde{\gamma}$ with $\tilde{\gamma}(t_j) = \gamma(t_j)$ will do.